A 7.55 g sample of an aqueous solution of
hydrochloric acid contains an unknown amount of
the acid.
If 29.1 mL of 0.321 M
sodium hydroxide are required to neutralize the
hydrochloric acid, what is the percent by mass of
hydrochloric acid in the mixture?
% by mass
Given: 29.1ml of 0.321M Sodium Hydroxide neutralizes 7.55g sample of Hydrochoric acid.
To Calculate : Percentage of Hydrochloric acid solution.
No of moles of Sodium Hydroxide present in 29.1ml of 0.321M solution = 0.321 x 29.1/1000 = 0.00934moles.
( No. of moles = Molarity x Volume expressed in litre.)
1 mole of sodium hydroxide neutralizes 1 mole of hydrochloric acid.
This means that 0.00934 moles of hydrocholric acid are present in the given solution.
Mass of 0.00934 moles of Hydrochloric acid = 0.00934 x 36.5 (Molar mass of HCl)
= 0.34091g of HCl.
Calculation of percent by mass of HCl:
= 4.545%
Ans : The percentage by mass of Hydrochloric acid in the geven solution is 4.545%.
A 7.55 g sample of an aqueous solution of hydrochloric acid contains an unknown amount of...
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